There are 49 mice in a pet shop.
30 mice are white.
27 mice are male.
18 mice have short tails.
8 mice are white and have short tails.
11 mice are male and have short tails.
7 mice are male but neither white nor short-tailed.
5 mice have all three characteristics and
2 have none.
Note:
W represents white mice
M represents male mice
A represents short-failed mice
I made a venn diagram but need help on the following:
1. n(M upside down u W)
2. n(M' U S)
3. P( both mice are short-tailed)
My teacher did not go over these notions so it would be helpful if you explain what it means and how you derive the numbers.....
The probability that Kate goes to the cinema is 0.2. If Kate does not go Claire goes. If Kate goes to the cinema the probability that she is late home is 0.3. If Clair goes to the cinema the probability that she is late home is 0.6.
Calculate the probability that
1. Kate goes to the cinema and is not late
2. The person who goes to the cinema arrives home late.
I will assume that you filled in the Venn diagrams correctly
n(M upside down u W) ---> n(M and W) = 14
n(M' U S) ---> n(M' or S)
Since S is the symbol used for the universal set, the count would be 49
P( both mice are short-tailed)
--- where does the "both" come from ???
------------------------------
Prob(kate goes to show) = .2
prob(Claire goes to show) = .8
prob(kate late) = .3 , prob(kate not late) = .7
prob(clair late) = .6 , prob(claire not late) = .4
prob(kate goes to show and it not late)
= .2(.7) = .14
prob(The person who goes to the cinema arrives home late)
= .2(.7) + .8(.4)
= .14 + .32
= .46
+54
1. To find n(M upside down u W), we need to calculate the number of mice that are both male and white. In the provided information, we are given that 8 mice are white and have short tails (n(A upside down u W) = 8). Also, 5 mice have all three characteristics (n(A upside down u W upside down u M) = 5). Since we know that 18 mice have short tails (n(A) = 18), we can subtract the number of mice that have all three characteristics to find n(M upside down u W):
n(M upside down u W) = n(A upside down u W) - n(A upside down u W upside down u M)
= 8 - 5
= 3
So there are 3 mice that are both male and white.
2. To find n(M' U S), we need to calculate the number of mice that are either not male or have short tails or both. In the provided information, we are given that 7 mice are male but neither white nor short-tailed (n(M ^ (W' U A')) = 7) and 2 mice have none of the three characteristics (n(W' ^ M' ^ A') = 2). So we can calculate n(M' U S) using the formula:
n(M' U S) = n(M ^ (W' U A')) + n(W' ^ M' ^ A')
= 7 + 2
= 9
So there are 9 mice that are either not male or have short tails or both.
3. To find the probability that both mice are short-tailed (P(both mice are short-tailed)), we need to consider the total number of mice (49) and the number of mice that have short tails (18). The probability can be calculated using:
P(both mice are short-tailed) = (n(A) / total number of mice) * ((n(A) - 1) / (total number of mice - 1))
= (18 / 49) * (17 / 48)
≈ 0.1092
So the probability that both mice are short-tailed is approximately 0.1092.
Note: The numbers for n(A upside down u W), n(A upside down u W upside down u M), n(M ^ (W' U A')), and n(W' ^ M' ^ A') were not explicitly given in the question. However, we can derive them based on the given information to calculate the desired quantities.
Now let's move on to the probability problems:
1. Probability that Kate goes to the cinema and is not late:
We know that the probability that Kate goes to the cinema is 0.2 (P(Kate goes to the cinema) = 0.2) and the probability that Kate is late if she goes to the cinema is 0.3 (P(Kate is late | Kate goes to the cinema) = 0.3). We want to calculate the probability that Kate goes to the cinema and is not late. We can use the multiplication rule for independent events to calculate this:
P(Kate goes to the cinema and is not late) = P(Kate goes to the cinema) * P(Kate is not late | Kate goes to the cinema)
= 0.2 * (1 - P(Kate is late | Kate goes to the cinema))
= 0.2 * (1 - 0.3)
= 0.2 * 0.7
= 0.14
So the probability that Kate goes to the cinema and is not late is 0.14.
2. Probability that the person who goes to the cinema arrives home late:
We have the probabilities for both Kate and Claire going to the cinema and being late if they go. We want to calculate the probability that the person who goes to the cinema arrives home late. We can use the law of total probability to calculate this:
P(The person who goes to the cinema arrives home late) = P(Kate goes to the cinema) * P(Kate is late | Kate goes to the cinema) + P(Claire goes to the cinema) * P(Claire is late | Claire goes to the cinema)
= 0.2 * 0.3 + (1 - 0.2) * 0.6
= 0.06 + 0.8 * 0.6
= 0.06 + 0.48
= 0.54
So the probability that the person who goes to the cinema arrives home late is 0.54.